Here we notice that $\ y=x$ in $G/G'$ and hence it is isomorphic to $\ C_7$.

We then proceed to obtain a presentation for $G'$ by only using the Schereier transversal of $\ C_7$.

However in another group, namely
$$\ G= \bigl\langle\, x, \ y \mid x^3 , \ y^3 , \ (xy)^3\,\bigr\rangle$$
we say that $G/G'$ is isomorphic to $\ C_3 \times \ C_3$ although by since the quotient is abelian, we find from the last relation that $\; y^3 = x^{-3} $, why don't we do the same?

Why in this second example we did not do as the first one although we have the same result that $\; y^3 = x^{-3}\; $ which means $\;y=x $.

1 Answer
1

Roots are not unique in groups: if $g^n=h^n$ it does not immediately follow that $g=h$. In your specific example, you have that $x^3=y^3=1$ - but the underlying group is precisely the group $C_3\times C_3$, one of the groups of order $9$. It is not cyclic of order $3$, and $x\neq y$. That is: that $x^3=y^{3}$ does not imply $x=y$. As another example, in $\mathbb{R}\setminus\{0\}$ under multiplication, $1^2=1$ and $(-1)^2=1$ but $1\neq -1$ [and you can mess around with the identity in any group - if $g\neq1$ and $g^n=1$ then $g^n=1^n$ but, by assumption, $g\neq 1$...].

More generally, if $g^n=h^n\neq1$ it does not immediately follow that $g=h$. For example, in the Quaternions we have that $i^2=j^2=k^2=-1$, but of course $i$, $j$ and $k$ are distinct elements. As another example, in $\mathbb{R}\setminus\{0\}$ under multiplication, $2^2=4$ and $(-2)^2=4$ but $2\neq-2$.)